60% . X + 1,5 .x = 1/6 . 2/3 +1/6 . 1/3
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15 tháng 1
a:
ĐKXĐ: x<>6
\(\dfrac{x-6}{-1,5}=\dfrac{-6}{x-6}\)
=>\(\left(x-6\right)^2=\left(-1,5\right)\cdot\left(-6\right)=9\)
=>\(\left[{}\begin{matrix}x-6=3\\x-6=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)
b: \(\dfrac{1\dfrac{2}{3}}{x-4}=\dfrac{5\dfrac{1}{6}}{x+1}\)
=>\(\dfrac{\dfrac{5}{3}}{x-4}=\dfrac{\dfrac{31}{6}}{x+1}\)
=>\(\dfrac{5}{3}\left(x+1\right)=\dfrac{31}{6}\left(x-4\right)\)
=>\(10\left(x+1\right)=31\left(x-4\right)\)
=>31x-124=10x+10
=>21x=134
=>\(x=\dfrac{134}{21}\)(nhận)
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0
3 tháng 7 2021
a) Ta có: \(2-x=2\left(x-2\right)^3\)
\(\Leftrightarrow-\left(x-2\right)-2\left(x-2\right)^3=0\)
\(\Leftrightarrow\left(x-2\right)\left[1+2\left(x-2\right)^2\right]=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
b) Ta có: \(8x^3-72x=0\)
\(\Leftrightarrow8x\left(x^2-9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy: S={0;3;-3}
c) Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)
\(\Leftrightarrow\left(x-1.5\right)^2\left[\left(x-1.5\right)^4+2\right]=0\)
\(\Leftrightarrow x-1.5=0\)
hay x=1,5
d) Ta có: \(2x^3+3x^2+3+2x=0\)
\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow2x=-3\)
hay \(x=-\dfrac{3}{2}\)
e) Ta có: \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
Vậy: S={0;1;-2}
f) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)
Vậy: S={0;2;12}
27 tháng 6 2019
1) \(|5x-3|=|7-x|\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=7-x\\5x-3=x-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x=10\\4x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)
Vậy...
27 tháng 6 2019
2) \(2.|3x-1|-3x=7\)
\(\Leftrightarrow2.|3x-1|=7+3x\)
\(\Leftrightarrow\orbr{\begin{cases}2.\left(3x-1\right)=7+3x\\2.\left(3x-1\right)=-7-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x-2=7+3x\\6x-2=-7-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=9\\9x=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{9}\end{cases}}\)
Vậy...
\(60\%x+1,5x=\frac{1}{6}.\frac{2}{3}+\frac{1}{6}.\frac{1}{3}\)
\(60\%x+1,5x=\frac{1}{6}\left(\frac{2}{3}+\frac{1}{3}\right)\)
\(60\%x+1,5x=\frac{1}{6}.1\)
\(60\%x+1,5x=\frac{1}{6}\)
\(\frac{60}{100}x+1,5x=\frac{1}{6}\)
\(\frac{3}{5}x+1,5x=\frac{1}{6}\)
\(x\left(\frac{3}{5}+1,5\right)=\frac{1}{6}\)
\(\frac{21}{10}.x=\frac{1}{6}\)
\(\Rightarrow x=\frac{5}{63}\)
\(60\%.x+1,5.x=\frac{1}{6}.\frac{2}{3}+\frac{1}{6}.\frac{1}{3}\)
\(\frac{60}{100}.x+1,5.x=\frac{1}{6}.\left(\frac{2}{3}+\frac{1}{3}\right)\)
\(\frac{3}{5}.x+\frac{15}{10}.x=\frac{1}{6}.\frac{3}{3}\)
\(x+\left(\frac{3}{5}.\frac{15}{10}\right)=\frac{1}{6}.1\)
\(x+\left(\frac{3}{1}.\frac{3}{10}\right)=\frac{1}{6}\)
\(x+\frac{9}{10}=\frac{1}{6}\)
\(x=\frac{1}{6}-\frac{9}{10}\)
\(x=\frac{10}{60}-\frac{54}{60}\)
\(x=\frac{-44}{60}\)
\(x=\frac{-11}{15}\)